Question

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

Short answer

a. The 6 probabilities are \( 1/9, 2/9, 1/9, 2/9, 1/9, 2/9 \)\nb. The probability of an odd number is \( 1/3 \), and of a number at most 3 is \( 4/9 \) \nc. The 6 probabilities are \( 1/21, 2/21, 3/21, 4/21, 5/21, 6/21 \). The probability of an odd number is \( 3/7 \), and of a number at most 3 is \( 2/7 \)

Step-by-step solution

Determine the value of \( p \)

The six simple events are \( O_{1} \), \( O_{2} \), \( O_{3} \), \( O_{4} \), \( O_{5} \), \( O_{6} \). Given that \( P(O_{1}) = p \), \( P(O_{2}) = 2p \), \( P(O_{3}) = p \), \( P(O_{4}) = 2p \), \( P(O_{5}) = p \), \( P(O_{6}) = 2p \), we use the fact that the sum of probabilities of all simple events is always equal to 1 to find \( p \). Adding these up, we get \( 9p = 1 \). Solving for \( p \) we find \( p = 1/9 \).

Calculate the probability of odd and at most 3

For an odd outcome, we sum the probabilities of \( O_{1} \), \( O_{3} \), and \( O_{5} \), giving us \( p + p + p = 3p \). Substituting the value of \( p \) from Step 1, we get \( 3/9 = 1/3 \). For an outcome of at most 3, we sum the probabilities of \( O_{1} \), \( O_{2} \), and \( O_{3} \), giving us \( p + 2p + p = 4p \). Substituting the value of \( p \) from Step 1, we get \( 4/9 \).

Determine the value of \( c \)

Here, the probability of each outcome is proportional to the number on the face, denoted as \( c, 2c, 3c, 4c, 5c, 6c \). Adding these up and equating to 1, we get \( 21c = 1 \). Solving for \( c \), we get \( c = 1/21 \).

Calculate new probabilities

For an odd outcome with the new dice, we sum the probabilities of \( O_{1} \), \( O_{3} \), and \( O_{5} \), giving us \( c + 3c + 5c = 9c = 9/21 = 3/7 \). For an outcome of at most 3 with the new dice, we sum the probabilities of \( O_{1} \), \( O_{2} \), and \( O_{3} \), giving us \( c + 2c + 3c = 6c = 6/21 = 2/7 \).